A.3 Solving Equations 1081 Notice that we need x ≥ 5 for this equation to have a solution. The radical is isolated. Square both sides. (This gets rid of the radical but may introduce extraneous roots. We’ll have to check our answers.) √ x + 1 2 = (x − 5) 2 x + 1 = x 2 − 10x + 25 We have obtained a quadratic. x 2 − 11x + 24 = 0 (x − 8)(x − 3) = 0 x = 8orx=3 Check: x = 8: √ 8 + 1 ? = 8 − 5 √ 9 ? = 3 3 = 3. Yes, x = 8 is a solution. Check: x = 3: √ 3 + 1 ? = 3 − 5 √ 4 ? = −2 2 =−2. No, x = 3isnot a solution. x = 8 is the only solution. ◆ As so often happens in science and mathematics, one question leads to another question. In this last example the question is, “Why did we get an answer that didn’t work?” Notice that if we didn’t check we wouldn’t have realized that x =3 did not work. To see what has gone on, we want to investigate squaring both sides with a simple example. Demonstration: x =−5 x 2 =(−5) 2 x 2 =25 x =± √ 25 =±5. Whoops. We’ve obtained an answer that doesn’t work in the original equation. The process of squaring both sides of an equation can give “answers” that don’t work. These are called extraneous roots. The reason we obtain an extraneous root in the problem is that the squaring process destroys sign information. The antidote is to check your answers. ◆ EXAMPLE A.34 Solve 2 √ x − √ x + 1 = 1. SOLUTION We can’t isolate both radicals simultaneously, so we isolate one of them. 2 √ x − 1 = √ x + 1 1082 APPENDIX A Algebra Square both sides to eliminate the radical on the right. (2 √ x − 1)(2 √ x − 1) = √ x + 1 2 4x − 2 √ x − 2 √ x + 1 = x + 1 4x − 4 √ x = x 3x = 4 √ x Square both sides again to eliminate the second radical. 9x 2 = 16x Now we have a quadratic equation. CAUTION We can’t divide by x because x may be zero. Instead, solve by factoring. 9x 2 − 16x = 0 x(9x −16) = 0 x = 0orx= 16 9 Check: x = 0: 2 √ 0 −1 ? = √ 0 +1 −1 = 1. x = 0isnot a solution. x = 16 9 : 2 16 9 − 1 ? = 16 9 + 1 2 4 3 − 1 ? = 16 9 + 9 9 8 3 − 1 ? = 25 9 8 3 − 3 3 ? = 5 3 5 3 = 5 3 . x = 16 9 is a solution. x = 16/9 is the only solution. ◆ MORAL When solving equations: You can multiply/divide both sides of the equation by anything except zero. A.3 Solving Equations 1083 If you multiply both sides of an equation by an expression that could be zero, you will introduce extraneous roots. This necessitates checking your answer(s). See Example A.26. If you divide both sides of an equation by an expression that could be zero you will lose that as a root. If you square both sides of an equation, 12 you may introduce extraneous roots. There- fore you must check your answers (either numerically or graphically). See Example A.33. Solving an Equation in x: Can you identify the equation as linear, quadratic, or radical in x? If it is linear in x, get all terms with x’s on one side; all else on the other. Factor out x and divide both sides by the coefficient of x. If it is quadratic in x, get everything on one side and zero on the other. Either use the quadratic formula or factor into two linear factors and set each factor equal to zero. If it is radical in x, isolate the radical and then eliminate it by squaring. Check for extraneous roots. If it is a higher degree polynomial, you may be lucky or you may not. Can you factor? Guess a root? Is it a quadratic in disguise? The number of roots is at most the degree of the equation. PROBLEMS FOR SECTION A.3 1. Find all y for which y 3 = y 2 . 2. Solve the equation λ 1 + x + 2 λ = 1 λβ for the variable indicated. (a) β (b) x (c) λ 3. Let f(x)=x 2 and g(x) = 1 + x. Find all w such that f g(w) − 2g(w) = 0. (You may find it simplest to solve for u = 1 + w and then find w.) 4. Solve for y. Find all y satisfying the equation. (a) y(2y − 3) = 5 (b) y(2y − 3) =−5 (c) y(y − 6) = 9 (d) y(y − 6) =−9 12 Raise both sides of an equation to any even power can introduce extraneous roots. 1084 APPENDIX A Algebra (e) y y−6 =−9 (f) 1 y−1 = 3 y 5. Let f(x)= √ x and g(x) = x +1. Solve the following equations. (a) f g(w) =g(w) (b) f(x 2 +1)−g(x +1) = 0 (c) 1 f(x) +3x = 2 f(x) 6. Solve for the indicated variables. (a) w(z +w) = z (z) (b) w(z +w) = z (w) (c) z+w w = z (z) (d) z+w w = z (w) (e) z 3 + 3z 2 + 2z = 0(z) (f) z 4 + 3z 2 + 2 = 0(z) (g) z 5 = 16z (z) B APPENDIX Geometric Formulas 1085 1086 APPENDIX B Geometric Formulas a bc consequently, we can calculate the distance between two points distance between (x 1 , y 1 ) and (x 2 , y 2 ) a 2 + b 2 = c 2 Pythagorean Theorem circumference = 2πr area = πr 2 volume = (Area of base) h = πr 2 h 2πr πrs + area of base Surface Area = 4πr 2 arclength s = rθ so s = rθ θ 2π arc length circumference |y 2 – y 1 | |x 2 – x 1 | (x 1 , y 1 ) (x 2 , y 2 ) (x 2 –x 1 ) 2 + (y 2 –y 1 ) 2 = = √ (∆x) 2 + (∆y) 2 √ r b b a c h s r b b h h s θ θ because = θ 2π = s 2πr area of a sector = r 2 θ θ 2π area of sector area of disk because = = θ 2π area of sector πr 2 ( ( ( ( 1 2 volume = πr 2 h right circular cone 1 3 1 3 volume = πr 3 4 3 Area = bh Area = bh Parallelogram: Area = bh Trapezoid: Area = ( ( 1 2 h b 2 h b 1 b 1 + b 2 2 h h h h h volume = abc Surface Area of closed box: add the areas of all 6 sides Prism volume = (Area of base)h volume = (Area of base)h Surface Area of closed cylinder: Add the areas of the base & lid to that of the sides. For the sides: unroll the cylinder Surface Area of closed cone: Remark: If you confuse the formulas for circumference of a circle, area of a disk, volume of a sphere, and surface area of a sphere, it may be useful to notice that the 1-dimensional measure has r 1 , the two-dimensional measure has r 2 , and the 3-dimensional measure has r 3 . r Figure B.1 f x ab f x ac b slope = f(b) – f(a) b– a C APPENDIX The Theoretical Basis of Applications of the Derivative In this section we will give the theoretical underpinnings of many of the statements we have been taking as fact without proof. We give proofs of statements 1 and 3–7 that follow. They have been numbered, as the proof of each rests on the veracity of previous statements. (Statements 5, 6, and 7 follow from the Mean Value Theorem (statement 4), whose proof rests on statement 3, whose proof depends on statements 1 and 2. Theorem 1 deals with an arbitrary function; Theorem 2 deals with continuous functions. Theorems 3 through 7 deal with functions that are (a) continuous on [a, b], and (b) differentiable on (a, b). These five theorems refer to functions whose graphs have no holes, no jumps, and no sharp corners. 1. Local Extremum Theorem. If a function f has a local maximum or local minimum at x = c, then either f (c) = 0orf (c) does not exist. 2. Extreme Value Theorem. If f is continuous on a closed interval [a, b], then f attains an absolute maximum and absolute minimum value on [a, b]. 3. Rolle’s Theorem. Suppose f is continuous on [a, b] and differentiable on (a, b). Then if f(a)= f(b),there exists some number c ∈ (a, b) such that f (c) = 0. 4. Mean Value Theorem. Suppose f is continuous on [a, b] and differentiable on (a, b). Then there exists some number c ∈ (a, b) such that f (c) = f(b)−f(a) b − a . 1087 1088 APPENDIX C The Theoretical Basis of Applications of the Derivative 5. Zero Derivative Theorem. Suppose f is continuous on [a, b] and differentiable on (a, b).Iff (x) = 0 for all x ∈ (a, b), then f is constant on (a, b). 6. Equal Derivatives Theorem. Suppose f is continuous on [a, b] and differentiable on (a, b).Iff (x) = g (x) for all x ∈ (a, b), then f(x) and g(x) differ by a constant on (a, b). That is, there exists a constant C such that f(x)=g(x) + C for all x ∈ (a, b). 7. Increasing / Decreasing Function Theorem. Suppose f is continuous on [a, b] and differentiable on (a, b). If f (x) > 0 for all x ∈ (a, b), then f is increasing on [a, b]; if f (x) < 0 for all x ∈ (a, b), then f is decreasing on [a, b]. Now let’s take up the job of proving these theorems. 1. Local Extremum Theorem: If a function f has a local maximum or local minimum at x = c, then either f (c) = 0orf (c) does not exist. c (i) cc (ii) c 1 c 2 (iii) c 1 c 2 (iv)(v) Figure C.1 Proof. Suppose that f has a local maximum at x = c. Then for all x sufficiently close to c, f(x)≤f(c). Therefore, for h sufficiently small, f(c+h) ≤ f(c) and f(c+h) − f(c)≤0. Now we want to divide both sides of the inequality by h for h sufficiently small but nonzero. We must treat the cases h>0and h<0separately; dividing by a positive number preserves the inequality while dividing by a negative number reverses it. 1 Case 1: h>0 f(c+h) − f(c) h ≤ 0 On both sides of the inequality compute the limit as h approaches zero from the right. lim h→0 + f(c+h) − f(c) h ≤ lim h→0 + 0 lim h→0 + f(c+h) − f(c) h ≤ 0 1 3 < 4but−3>−4. Multiplying (or dividing) both sides of an inequality by a negative number reverses the inequality. The Theoretical Basis of Applications of the Derivative 1089 Case 2: h<0 f(c+h) − f(c) h ≥ 0 On both sides of the inequality compute the limit as h approaches zero from the left. lim h→0 − f(c+h) − f(c) h ≥ lim h→0 − 0 lim h→0 − f(c+h) − f(c) h ≥ 0 If f (c) exists, then f (c) = lim h→0 + f(c+h) − f(c) h = lim h→0 − f(c+h) − f(c) h . f (c) ≥ 0 and f (c) ≤ 0, so we conclude that f (c) = 0iff (c) exists. We have shown that the Local Extremum Theorem is true if f has a local maximum at x = c. The argument in the case of a local minimum at x = c is analogous. REMARK We have seen that neither the condition that f (c) = 0 nor that f (c) is undefined is alone enough to guarantee a local extremum at x = c. 2. The Extreme Value Theorem is stated without proof. The proof is involved; please consult a more theoretical text for the proof. 3. Rolle’s Theorem: 2 Suppose f is continuous on [a, b] and differentiable on (a, b). Then if f(a)=f(b),there exists some number c ∈ (a, b) such that f (c) = 0. (See Figure C.2 below for illustrations. Basically, either f is constant on [a, b], in which case f (x) = 0on(a, b), or the graph of f must turn at least once in order to end up at the same height at which it began. At any turning point c, f (c) will be zero. (i)(ii) c 1 c c 2 (iii) f xx f ab x f (a, f(a)) (a, f(a)) (a, f(a))(b, f(b)) (b, f(b)) (b, f(b)) Figure C.2 Proof. Either f is constant on [a, b] or it is not. If f isconstant on [a, b], then f (x) = 0on(a, b) so c can be chosen to be anywhere on the interval (a, b). 2 Rolle’s Theorem is named after the French mathematician Michel Rolle, who published the result around 1697. 1090 APPENDIX C The Theoretical Basis of Applications of the Derivative Suppose f is not constant on [a, b]. The Extreme Value Theorem says that if f is continuous on a closed interval [a, b], then f attains an absolute maximum value and an absolute minimum value on [a, b]. Since f(a)=f(b),either the absolute maximum or the absolute minimum or both are attained on the open interval (a, b). Therefore there is some c ∈ (a, b) such that f has a local extremum at x = c. Since f is differentiable on (a, b), by the Local Extremum Theorem f (c) = 0. 4. Mean Value Theorem: If the function f is continuous on [a, b] and differentiable on (a, b), then there exists a number c ∈ (a, b) such that f (c) = f(b)−f(a) b − a . The Mean Value Theorem was first stated in the late 1700s by the French/Italian mathematician Joseph Louis Lagrange more than a century after Newton and Leibniz did the bulk of their work. 3 Geometrically, the Mean Value Theorem can be interpreted as telling us that between points A = (a, f(a))and B = (b, f(b))there is at least one point C = (c, f(c)) such that the slope of the tangent line through C is parallel to the secant line through A and B. See Figure C.3 below. (i)(ii) c 1 c 2 (iii) ab B C A f x c ab B A f x c 1 c 2 ab B A f x Figure C.3 Notice that Rolle’s Theorem is a special case of the Mean Value Theorem. For a practical interpretation, think of the word “mean” as the average. Then, if we let f(x)be a displacement function, the Mean Value Theorem can be interpreted as telling us that there must be at least one point in the interval (a, b) such that the instan- taneous velocity f (c) is the same as the average velocity on the interval [a, b]. This should strike you as quite reasonable; if a car averages 60 mph on the Massachusetts Turnpike (where its displacement function is both continuous and differentiable!), then at some instant the car must have been traveling at 60 mph. Proof. We prove the Mean Value Theorem by concocting a new function, v, from f such that the values of v at a and b are equal and Rolle’s Theorem can be applied to v. The function v is the vertical displacement function that gives the (signed) vertical displacement between a point (x, f(x))on the graph of f and the secant line through A = (a, f(a))and B = (b, f(b)). 3 It was Lagrange who introduced the prime notation for derivatives. . underpinnings of many of the statements we have been taking as fact without proof. We give proofs of statements 1 and 3–7 that follow. They have been numbered, as the proof of each rests on the veracity of. by the French/Italian mathematician Joseph Louis Lagrange more than a century after Newton and Leibniz did the bulk of their work. 3 Geometrically, the Mean Value Theorem can be interpreted as. or factor into two linear factors and set each factor equal to zero. If it is radical in x, isolate the radical and then eliminate it by squaring. Check for extraneous roots. If it is a higher